# Problem of the Week #64 - August 19th, 2013

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#### Chris L T521

##### Well-known member
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Here's this week's problem.

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Problem: Compute the cardinality of $\mathrm{SL}_3(\mathbb{F}_q)$, where $\mathbb{F}_q$ is a field of cardinality $q$.

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#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by Deveno. You can find his solution below.

I will choose to solve a somewhat easier problem: the cardinality of $$\displaystyle \ \text{GL}_3(\Bbb F_q)$$.

Since these are 3x3 matrices with non-zero determinant, their columns must be linearly independent. We can choose any non-zero vector in $$\displaystyle \Bbb (F_q)^3$$ as our first column, there are $$\displaystyle q^3 - 1$$ of these. For our second column, we can pick any non-zero vector that is not a scalar multiple of the first. This gives: $$\displaystyle q^3 -1 - (q - 1) = q^3 - q$$ choices for the second column, as there are $$\displaystyle q - 1$$ non-zero scalars to choose from. Finally, there are $$\displaystyle q^2 - 1$$ non-zero linear combinations of the first two chosen vectors, which we must exclude from our choice for the 3rd column, leaving: $$\displaystyle q^3 - 1 - (q^2 - 1) = q^3 - q^2$$ choices for column 3. Thus:

$$\displaystyle |\text{GL}_3(\Bbb F_q)| = (q^3-1)(q^3-q)(q^3-q^2)$$

Now we have the short exact sequence:

$$\displaystyle 0 \to \text{SL}_3(\Bbb F_q) \to \text{GL}_3(\Bbb F_q) \xrightarrow{det} (\Bbb F_q)^{\ast} \to 1$$

Which tells us that:

$$\displaystyle |\text{SL}_3(\Bbb F_q)| = \frac{|\text{GL}_3(\Bbb F_q)|}{|(\Bbb F_q)^{\ast}|} = \frac{(q^3-1)(q^3-q)(q^3-q^2)}{q-1} = q^3(q^2+q+1)(q-1)^2(q+1)$$

CLT Note: I just left my answer as $q^2(q^3-1)(q^3-q)$, but it doesn't really matter.

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